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%I #42 Dec 16 2019 10:23:30
%S 1,2,5,10,3,18,7,24,23,22,13,4,19,18,9,6,15,374,13,12,11,370,369,32,
%T 367,366,5,28,363,8,361,360,37,358,21,356,355,354,353,16,351,100,349,
%U 98,347,346,95,344,93,92,91,340,89,10,15,336,31,6,333,82,331,80,25,328,327,326
%N a(n) is the index of the first 0 term in the rumor sequence with initial 0th term 1 and parameters b = 2 and n.
%C A rumor sequence (running modulus recurrence sequence) is defined as follows: fix integer parameters b > 1 and n > 0. Set z[0] = any integer, and, for k > 0, define z[k] to be the least nonnegative residue of b*z[k-1] modulo (k+n). The rumor sequence conjecture states that all such rumor sequences are eventually 0.
%H B. Dearden, J. Iiams, and J. Metzger, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Dearden/dearden4.html">Rumor Arrays</a>, J. Integer Seq. 16 (2013), #13.9.3.
%H B. Dearden and J. Metzger, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Dearden/dearden3.html">Running Modulus Recursions</a>, J. Integer Seq. 13(1) (2010), #10.1.6.
%F a(n) = inf{m > 0 | z[0] = 1, z[m] = 0, and z[k] = (2*z[k-1] mod (k + n)) for k = 1..m}. - _Petros Hadjicostas_, Dec 13 2019
%e For n = 15, the z-sequence terms are 1, 2, 4, 8, 16, 12, 3, 6, 12, 0, so a(15) = 9; that is, z[0] = 1, z[1] = 2, z[2] = 4, ..., z[8] = 12, and z[9] = 0. [Edited by _Petros Hadjicostas_, Dec 13 2019]
%t For[n=1,n<50,n++,k=0;Clear[z];z[0]=1;z[k_]:=z[k]=Mod[2z[k-1],k+n];
%t While[z[k]>0,k++];Print[k];]
%Y Cf. A208125, A330411.
%K nonn,easy
%O 1,2
%A _Jerry Metzger_, Dec 10 2010
%E More terms from _Petros Hadjicostas_, Dec 13 2019
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